Question

The present age of Beeiia is 2 years more than Reeta. The present age of Teena is 3 years more than Beena If the sum of the ages of Reeta, Beena, and Teena is 79 years, find the present age of all the three of them. Solve the puzzle using the equations.

Answer 1

Dear Student,

Answer: Reeta's age , Beena's age and Teena's age 24 ,26 and 29 years respectively.

Solution:

Let the present age of Reeta is x years,then present age of Beena is x+2 years, so present age of Teena will be Beena's age+3 = x+2+3 = x+5

A.T.Q.

Reeta's age + Beena's age+ Teena's age = 79

x+x+2+x+5 =79

3x+7 = 79

3x = 79-7

3x= 72

x= 72/3

x= 24

So, Reeta age is 24 years

Beena's age is: 24+2 = 26 years

Teena's age is :26+3 = 29 years

let us check 24+26+29 =79

79 = 79

Hope it helps you.

Answer 2

Hi ,

At present :
________

Let Reeta age = x years

Beena age = 2 years more than Reeta

= ( x + 2 ) years

Teena age = 3 years more than Beena

= [ ( x + 2 ) + 3 ] years

= ( x + 5 ) years

according to the problem given ,

Sum of ages of Reeta , Beena and Teena

is 79 years.

x + ( x + 2 ) + ( x + 5 ) = 79

3x + 7 = 79

3x = 79 - 7

3x = 72

x = 72/3

x = 24

Therefore ,

Reeta age = x = 24 years

Beena's age = x + 2

= 24 + 2

= 26 years

Teena's age = x + 5

= 24 + 5

= 29 years

I hope this helps you.

: )